Integrate

1/(x^3+x^4)=A/x^3+B/x^2+C/x+D/x+1 ...
Comparing coefficients,A=1,B=-1,C=1,D=-1...
So, I=(1/x^3-1/x^2+1/x-1/(x+1))dx

=-(0.5)x^(-2)-x^(-1)+ln x -ln x-1+C

That's ok ..brutal but ok..I am looking for a shorter way..

\int \frac{x^{-4}}{\left ( \frac{1}{x} + 1\right )} dx

{\left ( \frac{1}{x} + 1\right )} = t\rightarrow -(x)^{-2} dx = dt

{\left ( \frac{1}{x} \right )}^2 = (t-1)^2

-\int \frac{(t-1)^2}{t} dt

-\int \frac{(t^2 -2t +1)}{t} dt

which can be done